---
title: "Mapping from s-plane to z-plane"
subject: "Digital Control System"
chapter: 3
key: "semester_6/digital_control_system:s-plane-to-z-plane-mapping"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/s-plane-to-z-plane-mapping
site: BCT Engineering
---

# Mapping from s-plane to z-plane

Digital Control System · Chapter 3: Analysis of Discrete Time Control System

- Asked in: 2073 Bhadra, 4 marks; 2071 Magh, 4 marks; 2070 Magh, 4 marks (asked 3 times)
- Video: [Mapping Between s plane and z plane](https://www.youtube.com/watch?v=RIkqWlFW7Bg) — Yasser Elshrief
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/s-plane-to-z-plane-mapping
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3

## Question

Explain one of the methods for mapping s-plane to z-plane. (How can s-plane be converted into z-plane and vice versa?)

## Short answer

Using z=e^{Ts} with s=σ+jω, magnitude |z|=e^{σT} and angle ∠z=ωT. The left half-plane maps inside the unit circle, right half-plane outside. The primary strip -ω_s/2<ω<ω_s/2 maps onto the entire z-plane, making the mapping many-to-one.

## Answer

The s-plane and z-plane are linked by the sampling relation $z = e^{Ts}$, where $T$ is the sampling period. The inverse relation is $s = \dfrac{1}{T}\ln z$.

#### Mapping $s \to z$

Put $s = \sigma + j\omega$:

$$
z = e^{T(\sigma + j\omega)} = e^{\sigma T}e^{j\omega T} \;\Rightarrow\; |z| = e^{\sigma T}, \quad \angle z = \omega T
$$

| s-plane | z-plane |
|---|---|
| Imaginary axis ($\sigma = 0$) | Unit circle $\lvert z\rvert = 1$ |
| Left half-plane ($\sigma < 0$) | Inside the unit circle |
| Right half-plane ($\sigma > 0$) | Outside the unit circle |
| Origin $s = 0$ | $z = 1$ |
| Constant $\sigma$ (vertical line) | Circle of radius $e^{\sigma T}$ |
| Constant $\omega$ (horizontal line) | Radial line at angle $\omega T$ |
| Constant $\zeta$ line | Logarithmic spiral |

As $\omega$ goes from $-\omega_s/2$ to $\omega_s/2$ (with $\omega_s = 2\pi/T$), the angle $\omega T$ goes from $-\pi$ to $\pi$, covering the circle once. This band is the **primary strip**. Each complementary strip above or below it maps onto the same z-plane again. So the mapping is many-to-one, and frequencies differing by $\omega_s$ cannot be told apart (aliasing).

```text
   s-plane (primary strip)       z-plane
   jw
   | ws/2 ---------             .---.
   |   LHP  | RHP   ==>       /  in  \  out
 --+--------+---> sigma      | (LHP)  |
   |        |                 \      /
   | -ws/2 ---------            '---'  |z|=1
```

#### Mapping $z \to s$

$$
s = \frac{1}{T}\ln z = \frac{1}{T}\left[\ln|z| + j\angle z\right]
$$

This gives the principal value in the primary strip.

**Example:** $T = 0.1$ s and $s = -2 + j10$. Then $z = e^{-0.2}\angle 1\text{ rad} = 0.819\angle57.3^\circ$. Going back: $\sigma = \ln(0.819)/0.1 = -2$ and $\omega = 1/0.1 = 10$ rad/s.

#### Use

The left half-plane maps to the inside of the unit circle. So a discrete system is stable when all its closed-loop poles lie inside $|z| = 1$. Specifications such as $\zeta$, $\omega_n$ and settling time can also be carried from the s-plane to the z-plane for design.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which relation maps the s-plane to the z-plane?
   A. z = sT
   B. z = e^(Ts)
   C. z = (1 + sT)/(1 - sT)
   D. z = ln(s)/T
   Answer: B. Sampling gives z = e^(Ts), with inverse s = (1/T) ln z. (Key point: Mapping relation)

2. Where does the left half of the s-plane (sigma < 0) map?
   A. Outside the unit circle
   B. On the unit circle
   C. Inside the unit circle
   D. To the negative real axis only
   Answer: C. |z| = e^(sigma T), which is less than 1 for sigma < 0. The imaginary axis maps to the unit circle. (Key point: Left half-plane maps inside unit circle)

3. Why is the s to z mapping many-to-one?
   A. Because |z| depends on omega
   B. Because the angle omega T repeats every 2 pi, so frequencies differing by ws map to the same z
   C. Because sigma has no effect on z
   D. Because z = 1 corresponds to every s
   Answer: B. The angle is omega T, so complementary strips of width ws all map onto the same z-plane, causing aliasing. (Key point: Primary strip and aliasing)

4. A horizontal line of constant omega in the s-plane maps to what in the z-plane?
   A. A circle of radius e^(omega T)
   B. A radial line at angle omega T
   C. A logarithmic spiral
   D. A single point
   Answer: B. Since the angle of z is omega T and its magnitude varies with sigma, a constant omega gives a radial line. (Key point: Constant omega lines)

## Related in Digital Control System

- [Mapping an s-plane region to the z-plane](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Amap-s-plane-region) — Digital Control System, chapter 3, asked 2×
- [Mapping constant damping ratio line to z-plane](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aconstant-damping-ratio-mapping) — Digital Control System, chapter 3

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